Consider the code snippet:
int main()
{
int n;
cout<<n;
float f;
cout.setf(ios_base::fixed,ios_base::floatfield);
cout<<endl<<f;
char ch;
cout<<endl<<static_cast<int>(ch);
return 0;
}
Output:
0
0.000000
0
Why it is not set to the garbage value instead of zero? Is it the consequence of undefined behavior? Compiled with g++4.8.2 on Ubuntu 14.04.
This is a compiler- and option-specific effect.
Static variables are zero-initialized first of all. Automatic local variables have no such guaranteed initialization. But a compiler can add zero-initialization, just to be “helpful”.
You're accidentally right that for the presented code the 0 result is “the consequence of undefined behavior”. But that's only because using an indeterminate value of type int is undefined behavior, or at least was UB up till and including C++11. It's a bit unclear whether accessing an indeterminate value of one of the three basic char types is UB: there are some Defect Reports about it (DR #240 and #616), and the C++14 wording is different from C++11 and earlier, and seems to have moved the specification of UB to some other part of the standard.
You should remember that there is no default initial value for automatic variables defined in the C++11 standard.
In practice, such a variable would contain some garbage value (if it sits in a processor register, the previous value of it; if it sits in a slot of the call stack, the former value at that location) which could be 0 (or something else).
Details are implementation specific and would vary with the compiler, the run, the optimization flags.
Read about undefined behavior, notably Lattner's blog: What every programmer should know about undefined behavior. Be really scared of them.
